Calculus Examples

$y = \sqrt{x + 7}$

Step 1

Find the first derivative.

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Step 1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 7}$ as ${(x + 7)}^{\frac{1}{2}}$ .

$f' (x) = \frac{d}{d x} ({(x + 7)}^{\frac{1}{2}})$

Step 1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x + 7$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $x + 7$ .

$f' (x) = \frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d x} (x + 7)$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{2}$ .

$f' (x) = \frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} (x + 7)$

Step 1.2.3

Replace all occurrences of $u$ with $x + 7$ .

$f' (x) = \frac{1}{2} \cdot {(x + 7)}^{\frac{1}{2} - 1} \frac{d}{d x} (x + 7)$

Step 1.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f' (x) = \frac{1}{2} \cdot {(x + 7)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x + 7)$

Step 1.4

Combine $- 1$ and $\frac{2}{2}$ .

$f' (x) = \frac{1}{2} \cdot {(x + 7)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x + 7)$

Step 1.5

Combine the numerators over the common denominator.

$f' (x) = \frac{1}{2} \cdot {(x + 7)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} (x + 7)$

Step 1.6

Simplify the numerator.

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Step 1.6.1

Multiply $- 1$ by $2$ .

$f' (x) = \frac{1}{2} \cdot {(x + 7)}^{\frac{1 - 2}{2}} \frac{d}{d x} (x + 7)$

Step 1.6.2

Subtract $2$ from $1$ .

$f' (x) = \frac{1}{2} \cdot {(x + 7)}^{\frac{- 1}{2}} \frac{d}{d x} (x + 7)$

Step 1.7

Combine fractions.

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Step 1.7.1

Move the negative in front of the fraction.

$f' (x) = \frac{1}{2} \cdot {(x + 7)}^{- \frac{1}{2}} \frac{d}{d x} (x + 7)$

Step 1.7.2

Combine $\frac{1}{2}$ and ${(x + 7)}^{- \frac{1}{2}}$ .

$f' (x) = \frac{{(x + 7)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d x} (x + 7)$

Step 1.7.3

Move ${(x + 7)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = \frac{1}{2 {(x + 7)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (x + 7)$

Step 1.8

By the Sum Rule, the derivative of $x + 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [7]$ .

$f' (x) = \frac{1}{2 {(x + 7)}^{\frac{1}{2}}} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (7))$

Step 1.9

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f' (x) = \frac{1}{2 {(x + 7)}^{\frac{1}{2}}} \cdot (1 + \frac{d}{d x} (7))$

Step 1.10

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$f' (x) = \frac{1}{2 {(x + 7)}^{\frac{1}{2}}} \cdot (1 + 0)$

Step 1.11

Simplify the expression.

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Step 1.11.1

Add $1$ and $0$ .

$f' (x) = \frac{1}{2 {(x + 7)}^{\frac{1}{2}}} \cdot 1$

Step 1.11.2

Multiply $\frac{1}{2 {(x + 7)}^{\frac{1}{2}}}$ by $1$ .

$f' (x) = \frac{1}{2 {(x + 7)}^{\frac{1}{2}}}$

Step 2

Find the second derivative.

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Step 2.1

Differentiate using the Constant Multiple Rule.

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Step 2.1.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2 {(x + 7)}^{\frac{1}{2}}}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [\frac{1}{{(x + 7)}^{\frac{1}{2}}}]$ .

$f'' (x) = \frac{1}{2} \cdot \frac{d}{d x} (\frac{1}{{(x + 7)}^{\frac{1}{2}}})$

Step 2.1.2

Apply basic rules of exponents.

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Step 2.1.2.1

Rewrite $\frac{1}{{(x + 7)}^{\frac{1}{2}}}$ as ${({(x + 7)}^{\frac{1}{2}})}^{- 1}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{d}{d x} ({({(x + 7)}^{\frac{1}{2}})}^{- 1})$

Step 2.1.2.2

Multiply the exponents in ${({(x + 7)}^{\frac{1}{2}})}^{- 1}$ .

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Step 2.1.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = \frac{1}{2} \cdot \frac{d}{d x} ({(x + 7)}^{\frac{1}{2} \cdot - 1})$

Step 2.1.2.2.2

Combine $\frac{1}{2}$ and $- 1$ .

$f'' (x) = \frac{1}{2} \cdot \frac{d}{d x} ({(x + 7)}^{\frac{- 1}{2}})$

Step 2.1.2.2.3

Move the negative in front of the fraction.

$f'' (x) = \frac{1}{2} \cdot \frac{d}{d x} ({(x + 7)}^{- \frac{1}{2}})$

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- \frac{1}{2}}$ and $g (x) = x + 7$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $x + 7$ .

$f'' (x) = \frac{1}{2} \cdot (\frac{d}{d u} (u^{- \frac{1}{2}}) \frac{d}{d x} (x + 7))$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - \frac{1}{2}$ .

$f'' (x) = \frac{1}{2} \cdot (- \frac{1}{2} u^{- \frac{1}{2} - 1} \frac{d}{d x} (x + 7))$

Step 2.2.3

Replace all occurrences of $u$ with $x + 7$ .

$f'' (x) = \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(x + 7)}^{- \frac{1}{2} - 1} \frac{d}{d x} (x + 7))$

Step 2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f'' (x) = \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(x + 7)}^{- \frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x + 7))$

Step 2.4

Combine $- 1$ and $\frac{2}{2}$ .

$f'' (x) = \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(x + 7)}^{- \frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x + 7))$

Step 2.5

Combine the numerators over the common denominator.

$f'' (x) = \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(x + 7)}^{\frac{- 1 - 1 \cdot 2}{2}} \frac{d}{d x} (x + 7))$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Multiply $- 1$ by $2$ .

$f'' (x) = \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(x + 7)}^{\frac{- 1 - 2}{2}} \frac{d}{d x} (x + 7))$

Step 2.6.2

Subtract $2$ from $- 1$ .

$f'' (x) = \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(x + 7)}^{\frac{- 3}{2}} \frac{d}{d x} (x + 7))$

Step 2.7

Combine fractions.

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Step 2.7.1

Move the negative in front of the fraction.

$f'' (x) = \frac{1}{2} \cdot (- \frac{1}{2} \cdot {(x + 7)}^{- \frac{3}{2}} \frac{d}{d x} (x + 7))$

Step 2.7.2

Combine ${(x + 7)}^{- \frac{3}{2}}$ and $\frac{1}{2}$ .

$f'' (x) = \frac{1}{2} \cdot (- \frac{{(x + 7)}^{- \frac{3}{2}}}{2} \cdot \frac{d}{d x} (x + 7))$

Step 2.7.3

Move ${(x + 7)}^{- \frac{3}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = \frac{1}{2} \cdot (- \frac{1}{2 {(x + 7)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (x + 7))$

Step 2.7.4

Multiply $\frac{1}{2}$ by $\frac{1}{2 {(x + 7)}^{\frac{3}{2}}}$ .

$f'' (x) = - \frac{1}{2 (2 {(x + 7)}^{\frac{3}{2}})} \cdot \frac{d}{d x} (x + 7)$

Step 2.7.5

Multiply $2$ by $2$ .

$f'' (x) = - \frac{1}{4 {(x + 7)}^{\frac{3}{2}}} \cdot \frac{d}{d x} (x + 7)$

Step 2.8

By the Sum Rule, the derivative of $x + 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [7]$ .

$f'' (x) = - \frac{1}{4 {(x + 7)}^{\frac{3}{2}}} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (7))$

Step 2.9

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - \frac{1}{4 {(x + 7)}^{\frac{3}{2}}} \cdot (1 + \frac{d}{d x} (7))$

Step 2.10

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$f'' (x) = - \frac{1}{4 {(x + 7)}^{\frac{3}{2}}} \cdot (1 + 0)$

Step 2.11

Simplify the expression.

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Step 2.11.1

Add $1$ and $0$ .

$f'' (x) = - \frac{1}{4 {(x + 7)}^{\frac{3}{2}}} \cdot 1$

Step 2.11.2

Multiply $- 1$ by $1$ .

$f'' (x) = - \frac{1}{4 {(x + 7)}^{\frac{3}{2}}}$

Step 3

Find the third derivative.

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Step 3.1

Differentiate using the Constant Multiple Rule.

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Step 3.1.1

Since $- \frac{1}{4}$ is constant with respect to $x$ , the derivative of $- \frac{1}{4 {(x + 7)}^{\frac{3}{2}}}$ with respect to $x$ is $- \frac{1}{4} \frac{d}{d x} [\frac{1}{{(x + 7)}^{\frac{3}{2}}}]$ .

$f''' (x) = - \frac{1}{4} \cdot \frac{d}{d x} \frac{1}{{(x + 7)}^{\frac{3}{2}}}$

Step 3.1.2

Apply basic rules of exponents.

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Step 3.1.2.1

Rewrite $\frac{1}{{(x + 7)}^{\frac{3}{2}}}$ as ${({(x + 7)}^{\frac{3}{2}})}^{- 1}$ .

$f''' (x) = - \frac{1}{4} \cdot \frac{d}{d x} {({(x + 7)}^{\frac{3}{2}})}^{- 1}$

Step 3.1.2.2

Multiply the exponents in ${({(x + 7)}^{\frac{3}{2}})}^{- 1}$ .

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Step 3.1.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f''' (x) = - \frac{1}{4} \cdot \frac{d}{d x} {(x + 7)}^{\frac{3}{2} \cdot - 1}$

Step 3.1.2.2.2

Multiply $\frac{3}{2} \cdot - 1$ .

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Step 3.1.2.2.2.1

Combine $\frac{3}{2}$ and $- 1$ .

$f''' (x) = - \frac{1}{4} \cdot \frac{d}{d x} {(x + 7)}^{\frac{3 \cdot - 1}{2}}$

Step 3.1.2.2.2.2

Multiply $3$ by $- 1$ .

$f''' (x) = - \frac{1}{4} \cdot \frac{d}{d x} {(x + 7)}^{\frac{- 3}{2}}$

Step 3.1.2.2.3

Move the negative in front of the fraction.

$f''' (x) = - \frac{1}{4} \cdot \frac{d}{d x} {(x + 7)}^{- \frac{3}{2}}$

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- \frac{3}{2}}$ and $g (x) = x + 7$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $x + 7$ .

$f''' (x) = - \frac{1}{4} \cdot (\frac{d}{d u} (u^{- \frac{3}{2}}) \frac{d}{d x} (x + 7))$

Step 3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - \frac{3}{2}$ .

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3}{2} u^{- \frac{3}{2} - 1} \frac{d}{d x} (x + 7))$

Step 3.2.3

Replace all occurrences of $u$ with $x + 7$ .

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3}{2} \cdot {(x + 7)}^{- \frac{3}{2} - 1} \frac{d}{d x} (x + 7))$

Step 3.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3}{2} \cdot {(x + 7)}^{- \frac{3}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x + 7))$

Step 3.4

Combine $- 1$ and $\frac{2}{2}$ .

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3}{2} \cdot {(x + 7)}^{- \frac{3}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x + 7))$

Step 3.5

Combine the numerators over the common denominator.

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3}{2} \cdot {(x + 7)}^{\frac{- 3 - 1 \cdot 2}{2}} \frac{d}{d x} (x + 7))$

Step 3.6

Simplify the numerator.

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Step 3.6.1

Multiply $- 1$ by $2$ .

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3}{2} \cdot {(x + 7)}^{\frac{- 3 - 2}{2}} \frac{d}{d x} (x + 7))$

Step 3.6.2

Subtract $2$ from $- 3$ .

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3}{2} \cdot {(x + 7)}^{\frac{- 5}{2}} \frac{d}{d x} (x + 7))$

Step 3.7

Combine fractions.

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Step 3.7.1

Move the negative in front of the fraction.

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3}{2} \cdot {(x + 7)}^{- \frac{5}{2}} \frac{d}{d x} (x + 7))$

Step 3.7.2

Combine ${(x + 7)}^{- \frac{5}{2}}$ and $\frac{3}{2}$ .

$f''' (x) = - \frac{1}{4} \cdot (- \frac{{(x + 7)}^{- \frac{5}{2}} \cdot 3}{2} \cdot \frac{d}{d x} (x + 7))$

Step 3.7.3

Simplify the expression.

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Step 3.7.3.1

Move $3$ to the left of ${(x + 7)}^{- \frac{5}{2}}$ .

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3 \cdot {(x + 7)}^{- \frac{5}{2}}}{2} \cdot \frac{d}{d x} (x + 7))$

Step 3.7.3.2

Move ${(x + 7)}^{- \frac{5}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f''' (x) = - \frac{1}{4} \cdot (- \frac{3}{2 {(x + 7)}^{\frac{5}{2}}} \cdot \frac{d}{d x} (x + 7))$

Step 3.7.3.3

Multiply $- 1$ by $- 1$ .

$f''' (x) = 1 (\frac{1}{4}) \cdot (\frac{3}{2 {(x + 7)}^{\frac{5}{2}}} \cdot \frac{d}{d x} (x + 7))$

Step 3.7.3.4

Multiply $\frac{1}{4}$ by $1$ .

$f''' (x) = \frac{1}{4} \cdot (\frac{3}{2 {(x + 7)}^{\frac{5}{2}}} \cdot \frac{d}{d x} (x + 7))$

Step 3.7.4

Multiply $\frac{3}{2 {(x + 7)}^{\frac{5}{2}}}$ by $\frac{1}{4}$ .

$f''' (x) = \frac{3}{2 {(x + 7)}^{\frac{5}{2}} \cdot 4} \cdot \frac{d}{d x} (x + 7)$

Step 3.7.5

Multiply $4$ by $2$ .

$f''' (x) = \frac{3}{8 {(x + 7)}^{\frac{5}{2}}} \cdot \frac{d}{d x} (x + 7)$

Step 3.8

By the Sum Rule, the derivative of $x + 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [7]$ .

$f''' (x) = \frac{3}{8 {(x + 7)}^{\frac{5}{2}}} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (7))$

Step 3.9

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f''' (x) = \frac{3}{8 {(x + 7)}^{\frac{5}{2}}} \cdot (1 + \frac{d}{d x} (7))$

Step 3.10

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$f''' (x) = \frac{3}{8 {(x + 7)}^{\frac{5}{2}}} \cdot (1 + 0)$

Step 3.11

Simplify the expression.

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Step 3.11.1

Add $1$ and $0$ .

$f''' (x) = \frac{3}{8 {(x + 7)}^{\frac{5}{2}}} \cdot 1$

Step 3.11.2

Multiply $\frac{3}{8 {(x + 7)}^{\frac{5}{2}}}$ by $1$ .

$f''' (x) = \frac{3}{8 {(x + 7)}^{\frac{5}{2}}}$

Step 4

Find the fourth derivative.

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Step 4.1

Differentiate using the Constant Multiple Rule.

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Step 4.1.1

Since $\frac{3}{8}$ is constant with respect to $x$ , the derivative of $\frac{3}{8 {(x + 7)}^{\frac{5}{2}}}$ with respect to $x$ is $\frac{3}{8} \frac{d}{d x} [\frac{1}{{(x + 7)}^{\frac{5}{2}}}]$ .

$f^{4} (x) = \frac{3}{8} \cdot \frac{d}{d x} (\frac{1}{{(x + 7)}^{\frac{5}{2}}})$

Step 4.1.2

Apply basic rules of exponents.

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Step 4.1.2.1

Rewrite $\frac{1}{{(x + 7)}^{\frac{5}{2}}}$ as ${({(x + 7)}^{\frac{5}{2}})}^{- 1}$ .

$f^{4} (x) = \frac{3}{8} \cdot \frac{d}{d x} ({({(x + 7)}^{\frac{5}{2}})}^{- 1})$

Step 4.1.2.2

Multiply the exponents in ${({(x + 7)}^{\frac{5}{2}})}^{- 1}$ .

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Step 4.1.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{4} (x) = \frac{3}{8} \cdot \frac{d}{d x} ({(x + 7)}^{\frac{5}{2} \cdot - 1})$

Step 4.1.2.2.2

Multiply $\frac{5}{2} \cdot - 1$ .

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Step 4.1.2.2.2.1

Combine $\frac{5}{2}$ and $- 1$ .

$f^{4} (x) = \frac{3}{8} \cdot \frac{d}{d x} ({(x + 7)}^{\frac{5 \cdot - 1}{2}})$

Step 4.1.2.2.2.2

Multiply $5$ by $- 1$ .

$f^{4} (x) = \frac{3}{8} \cdot \frac{d}{d x} ({(x + 7)}^{\frac{- 5}{2}})$

Step 4.1.2.2.3

Move the negative in front of the fraction.

$f^{4} (x) = \frac{3}{8} \cdot \frac{d}{d x} ({(x + 7)}^{- \frac{5}{2}})$

Step 4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- \frac{5}{2}}$ and $g (x) = x + 7$ .

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Step 4.2.1

To apply the Chain Rule, set $u$ as $x + 7$ .

$f^{4} (x) = \frac{3}{8} \cdot (\frac{d}{d u} (u^{- \frac{5}{2}}) \frac{d}{d x} (x + 7))$

Step 4.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - \frac{5}{2}$ .

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5}{2} u^{- \frac{5}{2} - 1} \frac{d}{d x} (x + 7))$

Step 4.2.3

Replace all occurrences of $u$ with $x + 7$ .

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5}{2} \cdot {(x + 7)}^{- \frac{5}{2} - 1} \frac{d}{d x} (x + 7))$

Step 4.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5}{2} \cdot {(x + 7)}^{- \frac{5}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x + 7))$

Step 4.4

Combine $- 1$ and $\frac{2}{2}$ .

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5}{2} \cdot {(x + 7)}^{- \frac{5}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x + 7))$

Step 4.5

Combine the numerators over the common denominator.

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5}{2} \cdot {(x + 7)}^{\frac{- 5 - 1 \cdot 2}{2}} \frac{d}{d x} (x + 7))$

Step 4.6

Simplify the numerator.

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Step 4.6.1

Multiply $- 1$ by $2$ .

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5}{2} \cdot {(x + 7)}^{\frac{- 5 - 2}{2}} \frac{d}{d x} (x + 7))$

Step 4.6.2

Subtract $2$ from $- 5$ .

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5}{2} \cdot {(x + 7)}^{\frac{- 7}{2}} \frac{d}{d x} (x + 7))$

Step 4.7

Combine fractions.

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Step 4.7.1

Move the negative in front of the fraction.

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5}{2} \cdot {(x + 7)}^{- \frac{7}{2}} \frac{d}{d x} (x + 7))$

Step 4.7.2

Combine ${(x + 7)}^{- \frac{7}{2}}$ and $\frac{5}{2}$ .

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{{(x + 7)}^{- \frac{7}{2}} \cdot 5}{2} \cdot \frac{d}{d x} (x + 7))$

Step 4.7.3

Simplify the expression.

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Step 4.7.3.1

Move $5$ to the left of ${(x + 7)}^{- \frac{7}{2}}$ .

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5 \cdot {(x + 7)}^{- \frac{7}{2}}}{2} \cdot \frac{d}{d x} (x + 7))$

Step 4.7.3.2

Move ${(x + 7)}^{- \frac{7}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f^{4} (x) = \frac{3}{8} \cdot (- \frac{5}{2 {(x + 7)}^{\frac{7}{2}}} \cdot \frac{d}{d x} (x + 7))$

Step 4.7.4

Multiply $\frac{3}{8}$ by $\frac{5}{2 {(x + 7)}^{\frac{7}{2}}}$ .

$f^{4} (x) = - \frac{3 \cdot 5}{8 (2 {(x + 7)}^{\frac{7}{2}})} \cdot \frac{d}{d x} (x + 7)$

Step 4.7.5

Multiply.

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Step 4.7.5.1

Multiply $3$ by $5$ .

$f^{4} (x) = - \frac{15}{8 (2 {(x + 7)}^{\frac{7}{2}})} \cdot \frac{d}{d x} (x + 7)$

Step 4.7.5.2

Multiply $2$ by $8$ .

$f^{4} (x) = - \frac{15}{16 {(x + 7)}^{\frac{7}{2}}} \cdot \frac{d}{d x} (x + 7)$

Step 4.8

By the Sum Rule, the derivative of $x + 7$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [7]$ .

$f^{4} (x) = - \frac{15}{16 {(x + 7)}^{\frac{7}{2}}} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (7))$

Step 4.9

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f^{4} (x) = - \frac{15}{16 {(x + 7)}^{\frac{7}{2}}} \cdot (1 + \frac{d}{d x} (7))$

Step 4.10

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$f^{4} (x) = - \frac{15}{16 {(x + 7)}^{\frac{7}{2}}} \cdot (1 + 0)$

Step 4.11

Simplify the expression.

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Step 4.11.1

Add $1$ and $0$ .

$f^{4} (x) = - \frac{15}{16 {(x + 7)}^{\frac{7}{2}}} \cdot 1$

Step 4.11.2

Multiply $- 1$ by $1$ .

$f^{4} (x) = - \frac{15}{16 {(x + 7)}^{\frac{7}{2}}}$